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klio [65]
3 years ago
12

An experiment was carried out at 25 °C by dissolving an unknown mass of magnesium ribbon into excess 1 mol dm–3 hydrochloric aci

d. The volume of gas produced over a period of 6 minutes was measured using a gas syringe. The result was plotted as shown in Figure 1
Time (min)

Figure 1 Volume of gas produced against time

a)Write a balanced equation for its reaction with hydrochloric acid.






(2 marks)
Use the data from Figure 1
to determine the total volume of gas produced (1mark)







to calculate the moles of gas produced at STP (2 marks)
[1 mole of any gas at STP has a volume of 22 400 cm3.]






to calculate the mass of limestone magnesium used.














(2 marks)

(i) What is meant by the term ‘rate of reaction’?


(1 mark )

Besides temperature, identify TWO other factors that can affect the rate of the reaction.




marks)
Chemistry
1 answer:
user100 [1]3 years ago
8 0

Answer:

to calculate the mass of limestone magnesium used.

Explanation:

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At a pressure of 9.25×10−14 atm and an ordinary temperature of 300.0 K , how many molecules are present in a volume of 1.10 cm3
statuscvo [17]

Answer:

2.49 × 10⁶ molecules

Explanation:

Given data

  • Pressure (P): 9.25 × 10⁻¹⁴ atm
  • Temperature (T): 300.0 K
  • Volume (V): 1.10cm^{3} .\frac{1mL}{1cm^{3}} .\frac{1L}{1000mL} =1.10\times 10^{-3} L

We can calculate the moles of gas using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 9.25 × 10⁻¹⁴ atm × 1.10 × 10⁻³ L / (0.0821 atm.L/mol.K) × 300.0 K

n = 4.13 × 10⁻¹⁸ mol

1 mole contains 6.02 × 10²³ molecules (Avogadro's number). The number of molecules in 4.13 × 10⁻¹⁸ moles is:

4.13 × 10⁻¹⁸ mol × (6.02 × 10²³ molecule/1 mol) = 2.49 × 10⁶ molecule

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Find [H+] of a 0.056 M hydrofluoric acid solution. Ka = 1.45 x 10-7
brilliants [131]

Answer:  [H^+] of 0.056 M HF solution is 8.96\times 10^{-5}

Explanation:

HF\rightarrow H^+F^-

 cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.056 M and \alpha = ?

K_a=1.45\times 10^{-7}

Putting in the values we get:

1.45\times 10^{-7}=\frac{(0.056\times \alpha)^2}{(0.056-0.056\times \alpha)}

(\alpha)=0.0016

[H^+]=c\times \alpha

[H^+]=0.056\times 0.0016=8.96\times 10^{-5}  

Thus [H^+] of 0.056 M HF solution is 8.96\times 10^{-5}

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